# Quickly Square Two, Three or Four digit numbers

In this post we will learn to mentally square two, three or four digits using an interesting trick in vedic maths.

We will use the concept of duplex of numbers to quickly square two digit numbers. Lets first understand what is a duplex of a number.

For a single digit number, say ‘a’ , Duplex of ‘a’ = $a^2$

For a number with two digits, say ‘ab’, Duplex of ‘ab’ = 2ab

Given below is the table containing the duplex of some numbers

Number | Duplex of the number |
---|---|

3 | Duplex of 3 => D(3) = $3^2$ = 9 |

9 | Duplex of 9 => D(9) = $9^2$ = 81 |

12 | Duplex of 12 => D(12) = 2 x 1 x 2 = 4 |

43 | Duplex of 43 => D(43) = 2 x 4 x 3 = 24 |

Now with this knowledge of Duplexes, we will see how we can find square of two digit numbers easily.

### Square of Two Digit Numbers

Consider a general two digit number, say, ‘ab’.

The square of ‘ab’ will have three parts.

ab^{2} = left most part| middle part | right most part

During calculations, we shall pass from the rightmost duplex to the leftmost duplex.

The rightmost part will be duplex of ‘b’, the middle part will be duplex of ‘ab’, finally the left most part will be duplex of ‘a’.

i.e ab^{2} = D(a)| D(ab) | D(b)

= a^{2} | 2ab | b$^2$

**Example 1:** Find the square of 12

12$^2$ = D(1)| D(12) | D(2)

= 1$^2$ | 2x1x2 | 2$^2$

= 1 | 4 | 4

Hence, 12 = 144 using the duplex methodology

**Example 2: **Find the square of 23

23$^2$ = D(2)| D(23) | D(3)

= 2$^2$ | 2x2x3 | 3$^2$

= 4 | 12 | 9

Here, the middle portion has more than two digits, **please note that only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.**

Hence for 12 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 4 to 5. Therefore,

23$^2$ = 4+1 | 2 | 9

Hence, 23$^2$ = 529 using the duplex methodology

### Square of Three Digit Numbers

Finding the square of three digit numbers is an extension of finding the squares of two digit numbers. In order to square 3 digit numbers using the duplex methodology we also need to find the duplex of 3 digit numbers in addition to the duplex of single and two digit numbers as described above.

For a number with three digits, say ‘abc’, Duplex of ‘abc’ => D(abc) = 2ac + $b^2$

For example,

- Duplex of 125 => D(125) = 2x1x 5 + $2^2$ = 14
- Duplex of 756 => D(756) = 2x7x6 + $5^2$ = 109

Now with this knowledge of Duplexes, we will see how we can square three digit numbers easily.

Consider a general three digit number, say, ‘abc’.

The square of ‘abc’ will have five parts as shown below(each part numbered with a digit for our convenience)

abc^{2} = 5 | 4 | 3 | 2 | 1

During calculations, we shall pass from the rightmost duplex to the leftmost duplex.

The rightmost part(1) will be duplex of ‘c’, the next part(2) will be duplex of bc, the middle part(3) will be duplex of ‘abc’, the next part(4) will be duplex of ab and finally the left most part(5) will be duplex of ‘a’.

i.e abc^{2 }= D(a) | D(ab) | D(abc)| D(bc) | D(c)

**Example 3:** Find the square of 321

321$^2$^{ }=D(3) | D(32) | D(321)| D(21) | D(1)

= 3$^2$ | 2x3x2 | 2x3x1 + 2$^2$ | 2x2x1 | 1$^2$

= 9 | 12 | 10 | 4 | 1

As mentioned above only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.

Hence for 10 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 12 to 13 and for 13 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 9 to 10.

Hence,

321$^2$ = 10 | 3 | 0 | 4 | 1

= 103041

**Example 4:** Find the square of 791

791^{2} =D(7) | D(79) | D(791)| D(91) | D(1)

= 7$^2$ | 2x7x9 | 2x7x1 + 9$^2$ | 2x9x1 | 1$^2$

= 49 | 126 | 95 | 18 | 1

= 625681

### Square of Four Digit Numbers

Proceeding on the same methodology as for 2 and 3 digit squares as mentioned above, to quickly square a 4 digit number, we must know duplex of 4 digit number as well.

For a number with four digits, say ‘abcd’, Duplex of ‘abcd’ => D(abcd) = 2ad + 2bc

The square of ‘abcd’ will have seven parts as shown below

abcd^{2} = D(a) | D(ab) | D(abc) | D(abcd)| D(bcd) | D(cd) | D(d)

= a2 | 2ab | 2ac + b2 | 2ad+2bc | 2bd + c2 | 2cd | d^{2}

**Example 5: **Find the square of 1221

1221$^2$= D(1) | D(12) | D(122) | D(1221)| D(221) | D(21) | D(1)

= 1$^2$ | 2x1x2 | 2x1x2 + 2$^2$ | 2x1x1+2x2x2 | 2x2x1 + 2$^2$ | 2x2x1 | 1$^2$

= 1 | 4 | 8 | 10 | 8 | 4 | 1

As mentioned above only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.

Hence for 10 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 8 to 9

1221$^2$ = 1 | 4 | 9 | 0 | 8 | 4 | 1

= 1490841

**Example 6: **Find the square of 9654

9654$^2$ = D(9) | D(96) | D(965) | D(9654)| D(654) | D(54) | D(4)

= 9$^2$ | 2x9x6 | 2x9x5 + 6$^2$ | 2x9x4+2x6x5 | 2x6x4 + 5$^2$ | 2x5x4 | 4$^2$

= 81 | 108 | 126 | 132 | 73 | 40 | 16

= 93199716

## Comments

## Vandana

Very helpful post Thank you so much!

## Vandana

I think there some thing missing after the text “For example, duplex of 125 =>” .

## Sujith Patnaik

Thanks Vandana. I missed to display the formula for a duplex of 3 digit number. Its updated now. Thank you 🙂

## REKHA

can u XPLAIN FOR A 4 DIGIT NUMBER WHICH HAS 0 IN IT . EG:1904

## Sujith Patnaik

I am extremely sorry for the delay Rekha.

1904$^2$ = D(1) | D(19) | D(190) | D(1904)| D(904) | D(04) | D(4)

= 1$^2$ | 2x1x9 | 2x1x0 + 9$^2$ | 2x1x4+2x9x0 | 2x9x4 + 0$^2$ | 2x0x4 | 4$^2$

= 1 | 18 | 81 | 8 | 72 | 0 | 16

Now, except for the left most part, excess digit from the other parts will be carried to the left. Hence, we have

= 3625216